Results 1 to 5 of 5

Math Help - Quadrilaterals and Pythagoras Theorem

  1. #1
    Newbie
    Joined
    Jun 2007
    Posts
    3

    Post Quadrilaterals and Pythagoras Theorem

    Hi, new to this,

    The diagonals AC and BD, of a quadrilateral ABCD, are perpendicular. Find the length of AB, if BC=5cm, DC=4cm and AD=3cm. Let the intersection of the diagonals be E.

    Now I'm thinking the answer will be in surd form. I've been thinking about the question for a while but i don't really know how to go about answering it.

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jun 2007
    From
    Cambridge, UK
    Posts
    41
    OK... here is a solution, using only Pythagoras' Theorem. The attached diagram shows your quadrilateral ABCD, with the diagonals intersecting at point E (all points labelled in blue capitals).

    Now, if the diagonals intersect at right angles, then all four of the angles labelled in red are right angles. So, the quadrilateral becomes a system of four right-angled triangles stuck together!

    Notice that I have labelled the lengths of four lines:

    BE = f; CE = g; DE = h; AE = j.

    Now, you just use Pythagoras on each one of the four triangles, separately. This gives you:

     f^2 + g^2 = (5 cm)^2
     g^2 + h^2 = (4 cm)^2
     h^2 + j^2 = (3 cm)^2
     j^2 + f^2 = x^2

    where x is the distance you want to measure. Now, do a little bit of algebra. Start with the first equation, subtract the second one, and then add the third one. You get the following:

    (f^2 + g^2) - (g^2 + h^2) + (h^2 + j^2) = f^2 + j^2 = x^2<br />

    and so

     x^2 = (5cm)^2 - (4cm)^2 + (3cm)^2 = (25 - 16 + 9)cm^2 =  18 cm^2
    x = \sqrt{18} cm \approx 4.24 cm
    Attached Thumbnails Attached Thumbnails Quadrilaterals and Pythagoras Theorem-sketch1.gif  
    Last edited by Pterid; June 4th 2007 at 07:24 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2007
    Posts
    3

    Post Thanks, want to clear something up.

    Thanks for the help,

    So is
    (f^2 + g^2) - (g^2 + h^2) + (h^2 + j^2) = f^2 + j^2 = x^2

    because that essentially is the same as
    f^2 + g^2 + h^2 + j^2 - g^2 - h^2 =  f^2 + j^2 = x^2

    The two g^2's and h^2's cancel out, equalling x^2

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2007
    From
    Cambridge, UK
    Posts
    41
    Yes, that's right.

    There is a better way to explain this, which another forum member suggested to me:

    Label the equations, 1 to 4:

    f^2 + g^2 = (5 cm)^2 (1)
    g^2 + h^2 = (4 cm)^2 (2)
    h^2 + j^2 = (3 cm)^2 (3)
    f^2 + j^2 = x^2 (4)


    Start with equation 1:

    f^2 + g^2 = (5 cm)^2

    "Subtract equation 2". This means, subtract the LHS of equation 2 from the LHS, and subtract the RHS of equation 2 from the RHS.

    f^2 + g^2 - g^2 - h^2 = (5 cm)^2 - (4 cm)^2
    f^2 - h^2 = (5 cm)^2 - (4 cm)^2

    (the h^2 cancels)

    Now "add equation 3" to this, just like before.

    f^2 - h^2 + h^2 + j^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2
    f^2 + j^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2

    The final step is to use equation 4: x^2 = f^2 + j^2

    so x^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2

    ...and there you have your answer!
    Last edited by Pterid; June 5th 2007 at 06:19 AM. Reason: cleanup
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2007
    Posts
    3

    Thanks

    Thanks loads. I think the 1st answer is easier to understand. 2 is always better than 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: February 11th 2011, 07:23 PM
  2. Ptolemy's Theorem and Cyclic Quadrilaterals
    Posted in the Geometry Forum
    Replies: 5
    Last Post: February 6th 2011, 04:28 PM
  3. Pythagoras' theorem
    Posted in the Geometry Forum
    Replies: 6
    Last Post: February 20th 2007, 12:00 PM
  4. Pythagoras' Theorem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 20th 2007, 02:41 AM
  5. Theorem on quadrilaterals
    Posted in the Geometry Forum
    Replies: 2
    Last Post: September 11th 2006, 10:01 AM

Search Tags


/mathhelpforum @mathhelpforum